Area under parabola

Problem:

You have a symmetric parabola shape like this, and you are trying to compute the area, what do you do?

Solution:

To generalize the problem let me denote $b$ as the base and $h$ as the height instead of those specific numbers

The first thing to do is to find the equation. It passes through $(-\frac{b}{2}, 0)$ and $(\frac{b}{2}, 0)$ suggest that it has the form $y = s(x - \frac{b}{2})(x + \frac{b}{2}) = s(x^2 - \frac{b^2}{4})$

It passes through $(0, h)$ suggest that $-s\frac{b^2}{4} = h$ or $s = -\frac{4h}{b^2}$.

So the full equation of the parabola is $-\frac{4h}{b^2}x^2 + h$.

The area can be found as

$\begin{eqnarray*} A &=& \int\limits_{-\frac{b}{2}}^{\frac{b}{2}}(-\frac{4h}{b^2}x^2 + h)dx \\ &=& -\frac{4h}{3b^2}x^3 + hx |_{-\frac{b}{2}}^{\frac{b}{2}} \\ &=& (-\frac{4h}{3b^2}(\frac{b}{2})^3 + h(\frac{b}{2})) - (-\frac{4h}{3b^2}(-\frac{b}{2})^3 + h(-\frac{b}{2})) \\ &=& (-\frac{hb}{6} + \frac{hb}{2}) - (\frac{hb}{6} - \frac{hb}{2}) \\ &=& \frac{2hb}{3} \end{eqnarray*}$

This last formula is really simple and worth remembering. A triangle is $\frac{hb}{2}$, and a parabola is $\frac{2hb}{3}$.